Below few options (BigQuery Standard SQL)
#standardSQL
WITH `yourproject.yourdataset.yourtable` AS (
SELECT 1 id, [STRUCT<num INT64, num_2 INT64>(2013, 2013),(1625, 1625),(1297, 1297),(7634, 7634)] nums UNION ALL
SELECT 2, [STRUCT<num INT64, num_2 INT64>(1,1),(2,2),(3,3),(4,4),(5,5)]
)
SELECT id, (SELECT SUM(num * num_2) FROM UNNEST(nums) nums ) as mul
FROM `yourproject.yourdataset.yourtable`
OR
#standardSQL
WITH `yourproject.yourdataset.yourtable` AS (
SELECT 1 id, [STRUCT<num INT64, num_2 INT64>(2013, 2013),(1625, 1625),(1297, 1297),(7634, 7634)] nums UNION ALL
SELECT 2, [STRUCT<num INT64, num_2 INT64>(1,1),(2,2),(3,3),(4,4),(5,5)]
)
SELECT id, SUM(num * num_2) AS mul
FROM `yourproject.yourdataset.yourtable`, UNNEST(nums) nums
GROUP BY id
both return below
id mul
1 66652959
2 55
If you would not need do SUM(num * num_2) but rather just have flattened your table - you can use below example
#standardSQL
WITH `yourproject.yourdataset.yourtable` AS (
SELECT 1 id, [STRUCT<num INT64, num_2 INT64>(2013, 2013),(1625, 1625),(1297, 1297),(7634, 7634)] nums UNION ALL
SELECT 2, [STRUCT<num INT64, num_2 INT64>(1,1),(2,2),(3,3),(4,4),(5,5)]
)
SELECT id, num, num_2
FROM `yourproject.yourdataset.yourtable`, UNNEST(nums) nums